Question

Consider the first-order initial value problem \(y^{\prime}(t)=a y+b, y(0)=A,\) for \(t \geq 0,\) where \(a, b,\) and \(A\) are real numbers. a. Explain why \(y=-b / a\) is an equilibrium solution and corresponds to a horizontal line in the direction field. b. Draw a representative direction field in the case that \(a>0\) Show that if \(A>-b / a,\) then the solution increases for \(t \geq 0\) and if \(A<-b / a,\) then the solution decreases for \(t \geq 0\). c. Draw a representative direction field in the case that \(a<0\) Show that if \(A>-b / a,\) then the solution decreases for \(t \geq 0\) and if \(A<-b / a,\) then the solution increases for \(t \geq 0\).

Short answer

#Answer#: (a) The equilibrium solution is \(y = -\frac{b}{a}\), and it is represented as a horizontal line in the direction field. (b) When \(a > 0\), if \(A > -\frac{b}{a}\), the solution \(y(t)\) will increase as \(t \geq 0\), and if \(A < -\frac{b}{a}\), the solution \(y(t)\) will decrease as \(t \geq 0\). (c) When \(a < 0\), if \(A > -\frac{b}{a}\), the solution \(y(t)\) will decrease as \(t \geq 0\), and if \(A < -\frac{b}{a}\), the solution \(y(t)\) will increase as \(t \geq 0\).

Step-by-step solution

(a) Finding the equilibrium solution and observing its horizontal line in the direction field

To find the equilibrium solution, set the derivative \(y^{\prime}(t)\) to zero and solve for \(y\). From the given differential equation, we have: \(y^{\prime}(t) = ay+b = 0\) Solving for \(y\), we get: \(y = -\frac{b}{a}\) This is the horizontal line of equilibrium where the solution stays constant. The direction field will have horizontal arrows indicating no change in \(y\) since the derivative is zero at the equilibrium solution.

(b) Analyzing the case where \(a>0\)

First, draw a representative direction field for the given differential equation when \(a > 0\). You will observe that the arrows representing positive values of \(a\) point upward (have positive slope) above the equilibrium line and downward (have negative slope) below the equilibrium line. Now, consider the given initial condition \(y(0) = A\). If \(A > -b/a\), then initially, the solution \(y(t)\) lies above the equilibrium line. Since the arrows for \(a>0\) in the direction field point upward above the equilibrium line, the solution will continue to increase as \(t \geq 0\). On the other hand, if \(A < -b/a\), then the solution \(y(t)\) lies below the equilibrium line. In this case, the arrows for \(a>0\) in the direction field point downward below the equilibrium line, implying that the solution will decrease as \(t \geq 0\).

(c) Analyzing the case where \(a

Now, draw a representative direction field for the given differential equation when \(a < 0\). You will observe that the arrows representing negative values of \(a\) point upward (have positive slope) below the equilibrium line, and downward (have negative slope) above the equilibrium line. Considering the given initial condition \(y(0) = A\), if \(A > -b/a\), then initially, the solution \(y(t)\) lies above the equilibrium line. Since the arrows for \(a<0\) in the direction field point downward above the equilibrium line, the solution will continue to decrease as \(t \geq 0\). On the other hand, if \(A < -b/a\), then the solution \(y(t)\) lies below the equilibrium line. In this case, the arrows for \(a<0\) in the direction field point upward below the equilibrium line, implying that the solution will increase as \(t \geq 0\).

Key concepts

Equilibrium Solution

An equilibrium solution in the context of differential equations refers to a solution where the rate of change is zero, meaning there is no growth or decline over time. For the given equation \(y^{\prime}(t)=ay+b\), setting \(y^{\prime}(t) = 0\) yields the equilibrium solution. Solving this, we find \(y = -\frac{b}{a}\). This solution represents a horizontal line in the direction field, indicating no change in \(y\). It's essential to understand that this line signifies the points where the system is at rest or balanced. This is why when the derivative of \(y\) is zero, the arrows in the direction field are horizontal, showing no slope anywhere along this equilibrium line.

Initial Value Problem

An initial value problem in differential equations sets the stage for determining a specific solution by providing an initial condition. For our problem, we're given \(y(0)=A\). This means at \(t = 0\), the solution must pass through the point \((0, A)\). The significance of the initial condition is that it helps in identifying one unique solution trajectory among possibly many solutions to the differential equation. In our context, the initial value \(A\) helps us see how the solution either increases or decreases over time depending on its position relative to the equilibrium point \(-\frac{b}{a}\). Whether \(a > 0\) or \(a < 0\) further influences the direction of change, determining if the slope of the solution will increase or decrease as time progresses.

First-Order Differential Equations

First-order differential equations involve derivatives of the first degree, meaning they contain terms like \(y^{\prime}(t)\) but not \(y^{\prime\prime}(t)\) or higher. These equations often model real-world scenarios where change depends on the current state, like in population growth or decay models. Our equation \(y^{\prime}(t)=ay+b\) is linear and first-order, indicating that the rate of change of \(y\) depends linearly on \(y\) itself.

• If \(a > 0\), any deviation from equilibrium tends to increase or decrease exponentially away or towards the equilibrium line \(-\frac{b}{a}\).
• If \(a < 0\), deviations will do the opposite, stabilizing toward equilibrium over time.

Understanding these concepts helps in visualizing and interpreting the direction fields and behavior of the solution over time.